我的活动中有一些碎片
[1], [2], [3], [4], [5], [6]
如果当前活动片段是[2],那么在返回按钮上按下我必须从[2]返回到[1],否则什么也不做。
最好的做法是什么?
编辑:应用程序不能从[3]…[6]返回[2]
我的活动中有一些碎片
[1], [2], [3], [4], [5], [6]
如果当前活动片段是[2],那么在返回按钮上按下我必须从[2]返回到[1],否则什么也不做。
最好的做法是什么?
编辑:应用程序不能从[3]…[6]返回[2]
当前回答
这是一个非常好的和可靠的解决方案:http://vinsol.com/blog/2014/10/01/handling-back-button-press-inside-fragments/
这家伙已经制作了一个抽象片段,处理backPress行为,并使用策略模式在活动片段之间切换。
对于你们中的一些人来说,抽象课程可能会有一些缺陷。
简单地说,链接中的解决方案是这样的:
// Abstract Fragment handling the back presses
public abstract class BackHandledFragment extends Fragment {
protected BackHandlerInterface backHandlerInterface;
public abstract String getTagText();
public abstract boolean onBackPressed();
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
if(!(getActivity() instanceof BackHandlerInterface)) {
throw new ClassCastException("Hosting activity must implement BackHandlerInterface");
} else {
backHandlerInterface = (BackHandlerInterface) getActivity();
}
}
@Override
public void onStart() {
super.onStart();
// Mark this fragment as the selected Fragment.
backHandlerInterface.setSelectedFragment(this);
}
public interface BackHandlerInterface {
public void setSelectedFragment(BackHandledFragment backHandledFragment);
}
}
和在活动中的用法:
// BASIC ACTIVITY CODE THAT LETS ITS FRAGMENT UTILIZE onBackPress EVENTS
// IN AN ADAPTIVE AND ORGANIZED PATTERN USING BackHandledFragment
public class TheActivity extends FragmentActivity implements BackHandlerInterface {
private BackHandledFragment selectedFragment;
@Override
public void onBackPressed() {
if(selectedFragment == null || !selectedFragment.onBackPressed()) {
// Selected fragment did not consume the back press event.
super.onBackPressed();
}
}
@Override
public void setSelectedFragment(BackHandledFragment selectedFragment) {
this.selectedFragment = selectedFragment;
}
}
其他回答
在您的活动中添加此代码
@Override
public void onBackPressed() {
if (getFragmentManager().getBackStackEntryCount() == 0) {
super.onBackPressed();
} else {
getFragmentManager().popBackStack();
}
}
并在commit()之前在Fragment中添加这一行
ft.addToBackStack(任何名称);
当你在Fragments之间转换时,调用addToBackStack()作为FragmentTransaction的一部分:
FragmentTransaction tx = fragmentManager.beginTransation();
tx.replace( R.id.fragment, new MyFragment() ).addToBackStack( "tag" ).commit();
如果你需要更详细的控制(例如,当一些片段可见时,你想要抑制返回键),你可以在你的片段的父视图上设置一个OnKeyListener:
//You need to add the following line for this solution to work; thanks skayred
fragment.getView().setFocusableInTouchMode(true);
fragment.getView().requestFocus();
fragment.getView().setOnKeyListener( new OnKeyListener()
{
@Override
public boolean onKey( View v, int keyCode, KeyEvent event )
{
if( keyCode == KeyEvent.KEYCODE_BACK )
{
return true;
}
return false;
}
} );
如果你正在使用FragmentActivity。然后像这样做
首先在你的碎片中调用这个。
public void callParentMethod(){
getActivity().onBackPressed();
}
然后调用你的父类FragmentActivity中的onBackPressed方法。
@Override
public void onBackPressed() {
//super.onBackPressed();
//create a dialog to ask yes no question whether or not the user wants to exit
...
}
我正在与SlidingMenu和Fragment一起工作,在这里展示我的案例,希望能帮助到别人。
按[后退]键时的逻辑:
When SlidingMenu shows, close it, no more things to do. Or when 2nd(or more) Fragment showing, slide back to previous Fragment, and no more things to do. SlidingMenu not shows, current Fragment is #0, do the original [Back] key does. public class Main extends SherlockFragmentActivity { private SlidingMenu menu=null; Constants.VP=new ViewPager(this); //Some stuff... @Override public void onBackPressed() { if(menu.isMenuShowing()) { menu.showContent(true); //Close SlidingMenu when menu showing return; } else { int page=Constants.VP.getCurrentItem(); if(page>0) { Constants.VP.setCurrentItem(page-1, true); //Show previous fragment until Fragment#0 return; } else {super.onBackPressed();} //If SlidingMenu is not showing and current Fragment is #0, do the original [Back] key does. In my case is exit from APP } } }
将addToBackStack()添加到片段事务中,然后使用下面的代码为片段实现反向导航
getSupportFragmentManager().addOnBackStackChangedListener(
new FragmentManager.OnBackStackChangedListener() {
public void onBackStackChanged() {
// Update your UI here.
}
});